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18 tháng 9 2021

\(x^2-16x-15x\left(x-4\right)\)

\(=x\left(x-4\right)\left(x+4\right)-15x\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2+4x-15x\right)\)

\(=\left(x-4\right)\left(x^2-11x\right)=x\left(x-4\right)\left(x-11\right)\)

18 tháng 9 2021

\(=x\left(x-4\right)\left(x+4\right)-15x\left(x-4\right)\\ =x\left(x-4\right)\left(x-11\right)\)

e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

5 tháng 8 2021

e)x3-4x+14x(x-2)=0

⇔ x(x2-4)+14x(x-2)=0

⇔ x(x-2)(x+2)+14x(x-2)=0

⇔ (x-2)(x2+2x+14x)=0

⇔ x(x-2)(x+16)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)

g)x2(x+1)-x(x+1)+x(x-1)=0

⇔ (x+1)(x2-x)+x(x-1)=0

⇔ x(x+1)(x-1)+x(x-1)=0

⇔ x(x-1)(x+2)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)

\(a^2x+a^2y+ax+ay+x+y\)

\(=a^2\left(x+y\right)+a\cdot\left(x+y\right)+\left(x+y\right)\)

\(=\left(x+y\right)\cdot\left(a^2+a+1\right)\)

14 tháng 9 2021

7. A = (x + y)^2 - 4y^2

= (x + y - 2y)(x + y + 2y)

= (x - y)(x + 3y)

14 tháng 9 2021

2. x^4 + 4

= x^4 + 4x^2 + 4 - 4x^2

= (x^2 + 2)^2 - (2x)^2

= (x^2 + 2x + 2)(x^2 - 2x + 2)

26 tháng 7 2021

bạn tách nhỏ câu hỏi ra

26 tháng 7 2021

19. 3x2-4x+1

= 3x2-3x-x+1

= (3x2-3x)-(x-1)

= 3x(x-1)-(x-1)

= (3x-1)(x-1)

20.3x2+4x-7

= 3x2+3x-7x-7

= (3x2+3x)-(7x+7)

= 3x(x+1)-7(x-1)

= (3x-7)(x-1)

21.3x2+7x-6

= 3x2+9x-2x-6

= (3x2+9x)-(2x+6)

= 3x(x+3)-2(x+3)

= (3x-2)(x+3)

22.3x2+3x-6

= 3x2+6x-3x-6

=(3x2+6x)-(3x+6)

= 3x(x+2)-3(x+2)

=(3x-3)(x+2)

= 3(x-1)(x+2)

23. 3x2-3x-6

=(3x2-6x)+(3x-6)

=3x(x-2)+3(x-2)

=(3x+3)(x-2)

= 3(x+1)(x-2)

24.6x2-13x+6

= 6x2-9x-4x+6

= (6x2-9x)-(4x-6)

=3x(2x-3)-2(2x-3)

=(3x-2)(2x-3)

25.6x2+13x+6

= 6x2+9x+4x+6

= (6x2+9x)+(4x+6)

=3x(2x+3)+2(2x+3)

=(3x+2)(2x+3)

26. 6x2+15x+6

= (6x2+12x)+(3x+6)

= 6x(x+2)+3(x+2)

=(6x+3)(x+2)

=3(2x+1)(x+2)

27. 6x2-15x+6

= (6x2-12x)-(3x-6)

= 6x(x-2)-3(x-2)

=(6x-3)(x-2)

=3(2x-1)(x-2)

28. 6x2+20x+6

= (6x2+18x)+(2x+6)

= 6x(x+3)+2(x+3)

= (6x+2)(x+3)

= 2(3x+1)(x+3)

29.6x2-20x+6

= (6x2-18x)-(2x-6)

= 6x(x-3)+2(x-3)

= (6x-2)(x-3)

= 2(3x-1)(x-3)

30.6x2+12x+6

= (6x2+6x)+(6x+6)

= 6x(x+1)+6(x+1)

= (6x+6)(x+1)

= 6(x+1)(x+1)

= 6(x+1)2

\(x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)

\(=\left(x-y\right)^2\left(x-y\right)-xy\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-2xy+y^2-xy\right)\)

\(=\left(x-y\right)\left(x^2-3xy+y^2\right)\)

5 tháng 8 2021

x(x-y)2 -y(x-y)2+xy2-x2y=x(x-y)2 -y(x-y)2+(xy2-x2y)=x(x-y)2 -y(x-y)2+xy(x-y)=\(\left(x-y\right)\left[x\left(x-y\right)-y\left(x-y\right)+xy\right]\)=\(\left(x-y\right)\left[\left(x-y\right)^2+xy\right]\)

5 tháng 8 2021

 2(x-y)2 -y(x-y)2 +xy2-x2y= 2(x-y)2-y(x-y)2+(xy^2-x^2y)=2(x-y)2-y(x-y)2+xy(x-y)=(x-y)\(\left[2\left(x-y\right)-y\left(x-y\right)+xy\right]\)=(x-y)(2x-2y-xy+y2+xy)=(x-y)(2x-2y+y2)

\(2\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)

\(=\left(x-y\right)^2\left(2-y\right)+xy\left(y-x\right)\)

\(=\left(x-y\right)^2\cdot\left(2-y\right)-xy\left(x-y\right)\)

\(=\left(x-y\right)\left[\left(x-y\right)\left(2-y\right)-xy\right]\)

o: x^4+x^3+x^2-1

=x^3(x+1)+(x-1)(x+1)

=(x+1)(x^3+x-1)

q: \(=\left(x^3-y^3\right)+xy\left(x-y\right)\)

=(x-y)(x^2+xy+y^2)+xy(x-y)

=(x-y)(x^2+2xy+y^2)

=(x-y)(x+y)^2

s: =(2xy)^2-(x^2+y^2-1)^2

=(2xy-x^2-y^2+1)(2xy+x^2+y^2-1)

=[1-(x^2-2xy+y^2]+[(x+y)^2-1]

=(1-x+y)(1+x-y)(x+y-1)(x+y+1)

u: =(x^2-y^2)-4(x+y)

=(x+y)(x-y)-4(x+y)

=(x+y)(x-y-4)

x: =(x^3-y^3)-(3x-3y)

=(x-y)(x^2+xy+y^2)-3(x-y)

=(x-y)(x^2+xy+y^2-3)

z: =3(x-y)+(x^2-2xy+y^2)

=3(x-y)+(x-y)^2

=(x-y)(x-y+3)

29 tháng 8 2023

o) \(x^4+x^3+x^2-1\)

\(=\left(x^4+x^3\right)+\left(x^2-1\right)\)

\(=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)\)

\(=\left(x+1\right)\left(x^3+x-1\right)\)

q) \(x^3+x^2y-xy^2-y^3\)

\(=\left(x^3+x^2y\right)-\left(xy^2+y^3\right)\)

\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)\)

\(=\left(x+y\right)^2\left(x-y\right)\)

s) \(4x^2y^2-\left(x^2+y^2-1\right)^2\)

\(=\left(2xy\right)^2-\left(x^2+y^2-1\right)^2\)

\(=\left(2xy-x^2-y^2+1\right)\left(2xy+x^2+y^2-1\right)\)

\(=-\left(x^2-2xy+y^2-1\right)\left(x^2+2xy+y^2-1\right)\)

\(=-\left(x-y-1\right)\left(x-y+1\right)\left(x+y+1\right)\left(x+y-1\right)\)

u) \(x^2-y^2-4x-4y\)

\(=\left(x^2-y^2\right)-\left(4x+4y\right)\)

\(=\left(x+y\right)\left(x-y\right)-4\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-4\right)\)

x) \(x^3-y^3-3x+3y\)

\(=\left(x^3-y^3\right)-\left(3x-3y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2-3\right)\)

z) \(3x-3y+x^2-2xy+y^2\)

\(=\left(3x-3y\right)+\left(x^2-2xy+y^2\right)\)

\(=3\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3+x-y\right)\)

AH
Akai Haruma
Giáo viên
15 tháng 10 2021

Lời giải:

$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$

$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$

20 tháng 8 2021

\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)

Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)

\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)